来自满足给定条件的数组中的对

给定数组arr [] ，任务是计算数组中所有有效对。如果func(arr [i])+ func(arr [j])= func(XOR(arr [i]，arr [j]))，则说成对(arr [i]，arr [j])是有效的其中FUNC(x)返回比特集合x中的数目。

例子：

Input: arr[] = {2, 3, 4, 5, 6} 
Output: 3 
(2, 4), (2, 5) and (3, 4) are the only valid pairs.

Input: arr[] = {12, 13, 34, 25, 6} 
Output: 4 
 

方法：迭代每个可能的对，并检查该对是否满足给定条件。如果满足条件，则更新计数=计数+ 1 。最后打印计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the number
// of set bits in n
int setBits(int n)
{
    int count = 0;
 
    while (n) {
        n = n & (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count of required pairs
int countPairs(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n - 1; i++) {
 
        // Set bits for first element of the pair
        int setbits_x = setBits(a[i]);
 
        for (int j = i + 1; j < n; j++) {
 
            // Set bits for second element of the pair
            int setbits_y = setBits(a[j]);
 
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
 
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
 
                // Increment the count
                count++;
        }
    }
 
    // Return the total count
    return count;
}
 
// Driver code
int main()
{
    int a[] = { 2, 3, 4, 5, 6 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << countPairs(a, n);
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function to return the number
// of set bits in n
static int setBits(int n)
{
    int count = 0;
 
    while (n > 0)
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count of
// required pairs
static int countPairs(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n - 1; i++)
    {
 
        // Set bits for first element
        // of the pair
        int setbits_x = setBits(a[i]);
 
        for (int j = i + 1; j < n; j++)
        {
 
            // Set bits for second element
            // of the pair
            int setbits_y = setBits(a[j]);
 
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
 
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
 
                // Increment the count
                count++;
        }
    }
 
    // Return the total count
    return count;
}
 
    // Driver code
    public static void main (String[] args)
    {
 
        int []a = { 2, 3, 4, 5, 6 };
        int n = a.length;
        System.out.println(countPairs(a, n));
    }
}
 
// This code is contributed by ajit.

Python3

# Python 3 implementation of the approach
 
# Function to return the number
# of set bits in n
def setBits(n):
    count = 0
 
    while (n):
        n = n & (n - 1)
        count += 1
 
    return count
 
# Function to return the count
# of required pairs
def countPairs(a, n):
    count = 0
 
    for i in range(0, n - 1, 1):
         
        # Set bits for first element
        # of the pair
        setbits_x = setBits(a[i])
 
        for j in range(i + 1, n, 1):
             
            # Set bits for second element
            # of the pair
            setbits_y = setBits(a[j])
 
            # Set bits of the resultant number
            # which is the XOR of both the
            # elements of the pair
            setbits_xor_xy = setBits(a[i] ^ a[j]);
 
            # If the condition is satisfied
            if (setbits_x +
                setbits_y == setbits_xor_xy):
                 
                # Increment the count
                count += 1
 
    # Return the total count
    return count
 
# Driver code
if __name__ == '__main__':
    a = [2, 3, 4, 5, 6]
 
    n = len(a)
    print(countPairs(a, n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# implementation of the approach
using System;
class GFG
{
 
// Function to return the number
// of set bits in n
static int setBits(int n)
{
    int count = 0;
 
    while (n > 0)
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count of
// required pairs
static int countPairs(int []a, int n)
{
    int count = 0;
 
    for (int i = 0; i < n - 1; i++)
    {
 
        // Set bits for first element
        // of the pair
        int setbits_x = setBits(a[i]);
 
        for (int j = i + 1; j < n; j++)
        {
 
            // Set bits for second element
            // of the pair
            int setbits_y = setBits(a[j]);
 
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
 
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
 
                // Increment the count
                count++;
        }
    }
 
    // Return the total count
    return count;
}
 
// Driver code
public static void Main()
{
    int []a = { 2, 3, 4, 5, 6 };
 
    int n = a.Length;
 
    Console.Write(countPairs(a, n));
}
}
 
// This code is contributed
// by Akanksha Rai

Javascript

 
// Javascript implementation of the approach
     
// Function to return the number
// of set bits in n
function setBits(n)
{
    let count = 0;
   
    while (n > 0)
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count of
// required pairs
function countPairs(a, n)
{
    let count = 0;
   
    for(let i = 0; i < n - 1; i++)
    {
         
        // Set bits for first element
        // of the pair
        let setbits_x = setBits(a[i]);
   
        for(let j = i + 1; j < n; j++)
        {
   
            // Set bits for second element
            // of the pair
            let setbits_y = setBits(a[j]);
   
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            let setbits_xor_xy = setBits(a[i] ^ a[j]);
   
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
   
                // Increment the count
                count++;
        }
    }
   
    // Return the total count
    return count;
}
 
// Driver code
let a = [ 2, 3, 4, 5, 6 ];
let n = a.length;
 
document.write(countPairs(a, n));
 
// This code is contributed by unknown2108
 

输出：

3